3. Mixtures 2014

1

3. Mixtures

2014

3.1 Quanties of mixtures

A mixture is system where any phase containing more than one component.

A mixture may be gas, liquid or solid.

In mixing, extensive quantities are seldom additive. For example, if 1 litre sulphuric acid is mixed with 1 litre water, the volume of the

mixture will be about 1.8 litre.

Mass is always additive. V, H are additive in ideal mixtures (see later, equation 3.8). S, A, G are never additive,because entropy of mixing is positive (see subsection 3.8).

3

Mixtures can be characterised by the deviation from additivity. (We define these quantities for two component mixtures). Example:

Volume of mixing (the change of volume in mixing):





mix

V  V  n V  n V



volume of mixture molar volumes of pure components





mix

E  E  n E  n E



In general: let E any extensive quantity (H, S, G, A , etc):

The next definitions are valid for isothermal-isobaric processes, i.e. T and p are the same after mixing as

before, see Subsection 2.2.

(3.1)

(3.2)





* 1 1

2 1

2 1

m mix m

m

mix

x V x V

n n

V n

n

V V  

 



 



For one mole of mixture - molar volume of mixing (x mole fraction)

The reason of the mixing is the

intermolecular

interaction, i.e. the molecules form associates, e.g.

hydrogen bonds.

_mixV_m

(cm³/mol)

x



-1

0 1

(3.2)

Fig. 3.1, mixture: water – ethanol

5





* 1 1

2 1

m m

m

mix

x H x H

n n

H H  

 



Molar enthalpy of mixing (division with n=n₁+n₂)

molar enthalpy of solution

molar enthalpies of pure components

If _mixH_m > 0, endothermic - we must add heat to the system to keep the temperature unchanged

If _mixH_m < 0, exothermic - heat is given away by the system.

Weight fractions (mass fractions) are frequently used in techical diagrams.

(3.4)

 ^mix h (kJ/kg)

w(ethanol)

0 1

-50 0

The specific enthalpy of mixing of water-ethanol system at three temperatures (w: weight fraction):

0 ^oC 50 ^oC

80 ^oC

Fig. 3.2

3.2 Intermolecular interactions

They maybe

- electrostatic, e.g. benzene-toluene mixture; - dipole-dipole, e.g. acetone-thiophene;

- hydrogen bond, e.g. ethanol-water.

It is possible also that only molecules of one component build associates (selfassociation), and the interaction with the molecules of the other component(s) is weak.

The ethanol-water liquid mixture, models

*This atoms or electron pairs can participate in H-bonds

Fig. 3.3a

Fig. 3.3b

Pay attention!

The solution is a mixture. The mixture is called solution if one component is the bulk of the mixture, this is the solvent. All other components in smaller concentrations are the solutes.

Gibbs free energy of mixing: _mixG

In a spontaneous process at constant temperature and pressure G decreases, (2.57) .

Two components are miscible if the Gibbs function of mixing is negative.

_mixG_m = _mixH_m - T _mixS_m

may be

negative or positive

always positive

H and S depend on T, too.

(3.5)

The mixture building is a spontaneous

process. Therefore the entropy will be increase during mixing.

Remember: S=k*lnW ⁽Subsection 1.16 and Equ.1.80)

k: Boltzmann constant

W: Thermodynamic probability: number of

microstates belonging to system with N atom.

Microstate: a possible distribution of particles under the energy levels of the system.

Extensive quantities have partial molar values.

First we discuss partial molar volumes.

If we add one mol (18 cm³) water to very much water, the volume will increase by 18 cm³.

If we add one mol (18 cm³) water to very much ethanol, the volume will increase by 14 cm³ only.

Explanation: water molecules surrounded by ethanol molecules occupy different volumes than water

molecules surrounded by water molecules, see figures 3.3.

3.3. Partial molar quantities

We say that the partial molar volume of water is 18 cm³/mol in water and

14 cm³/mol in ethanol.

The partial molar volume is the function of concentration, function of the intermolecular interactions in the mixture.

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The definition of partial molar volumes (in a two component system)

, 2

1 , 1

n T

n

V V 



 







 

, 1

2 , 2

n T

n

V V 



 







 

The partial molar volume of a component is the change of volume of the mixture if one mole of a

component is added to infinite amount of mixture at constant temperature and pressure.

Infinite: so that the composition (theoretically) does not change (subsection 2.9).

(3.5)

16

H₂O ethanol

14 16 18

x(ethanol) 

0 1

54 56 58 V_w

(cm³/mol)

V_e

(cm³/mol )

Water-ethanol system

Fig. 3.5

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At constant T and p the volume of a two component system depends on the amounts of components only:

V = V(n₁, n₂) Complete differential:

2 ,

2 , 1

,

1 , _{2 1}

n dn dn V

n dV V

n T p n

T

p 

 







 



 







 

2 2

1

1

dn V dn

V

dV  

V = V₁n₁ +V₂n₂

(3.6)

Fig. 3.6

(3.7)

The volume of the mixture equals the number of moles of A times the partial volume of A, plus the number of moles of B times the partial volume of B.

(It is valid both for ideal and for real solutions.) In ideal solution:

2

* 2 1

* 1 2

* 2 1

*

1

n V n V V V V

V

V 

 







In an ideal mixture the partial molar volume is

equal to the molar volume of pure component (the

‘*’ superscript refers to the pure component).

(3.8)

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Other extensive parameters (H, G, etc.) also have partial molar quantities.

Denote the extensive quantity by E i

n j E E

nj

T i p

i  



 







 

, ,

The partial molar value of an extensive quantity is the change of that quantity if one mole of the

component is added to infinite amount of mixture at constant temperature and pressure.

(3.9)

2 2

1

1dn E dn

E

dE  

In a two component system:

E = E₁n₁ +E₂n₂

The extensive quantity of the mixture is the sum of partial molar quantities times the amounts in moles. In a multicomponent system:

i i

dn E

dE ^  ^{E }  ^E

ⁿ

(3.10a)

(3.10b)

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The partial molar Gibbs function is chemical potential

nj

p i T

i

n

G

, ,

 

 







 



For a two component system at constant T and p:

2 2

1 1

,

dn dn

dG

   

2 2 1

1

n n

G    

The Gibbs function of the mixture is the sum of chemical potentials times the amounts in moles.

(3.11)

(3.12a) (3.12b)

2 2 1

1

n n

G    

2 2

1 1

,

dn dn

dG

   

The Gibbs-Duham equation

We derive it for chemical potentials but it is valid for other partial molar quantities, too. Equations (2.44)!

The complete differential (at constant p and T) ^(2.44b)

2 2

2 2

1 1

1 1

,

 dn n d   dn n d 

dG

   

2

0

2 1

1

d   n d   n

Subtracting the (2.44a) equation from this one:

(3.13) is the Gibbs-Duham equation (it is valid when T and p does not change (only the composition changes).

(2.44a⁾

(3.13)

23

I.e., the chemical potentials of the two components are not independent. (If we know the dependence of

₁ on the composition, we can calculate that of ₂.)

2

0

2 1

1

d   n d   n

Since n₁ and n₂ are always positive, if ₁

increases, ₂ decreases, and the other way round.

Where one of them has a maximum (d₁= 0), the other one has a minimum (d₂ = 0, too).

2

0

2 1

1

dV  n dV  n

Gibbs-Duhem like equations are valid also for other extensive properites, for volumes:

We can interpret the partial molar volume diagram of the water-ethanol system (see Fig. 3.5).

(3.14)

3.4 Determination of partial molar quantities

We discuss two methods. Example: partial molar volume.

, 1

2 , 2

n T

n

V V 



 







 

1. Method of slopes

(3.15)

25

The mole fraction and the slope have to be determined at several points of the curve. We

obtains V₂-n₂ data pairs.

V

n₂ n₁=const



n1

, T , 2 p

2

n

tan V

V 



 







 

 

Fig. 3.7

2. The method of intercepts

2 2

1

1dn V dn

V

dV  

2 2 1

1n V n

V

V  

Dividing by (n₁+n₂)

2 2 1

1x V x

V

V_m  

2 2

1

1dx V dx

V

dV_m   x₁ = 1-x₂ dx₁ = -dx₂

2 2 2

1(1 x ) V x V

V_m   

2 1

2

1 (V V )x

V

V_m   

2 1

2 )

(V V dx dV_m  

)

( _{2 1}

2

V dx V

dV_m





2

1 x

dx V dV

V_m   ^{m (3.16)}

27

2 2

1 x

dx V dV

V_m   ^m This is the equation of the tangent of the V_m-x₂ curve.

x₂ 0

V_m

* 1

V

* 2

V

x₂ 

1

)

( _{2 1}

2

V dx V

dV_m



 V₁

V₂

Fig. 3.8

The intercepts of the tangents of the V_m-x₂ curve produce the partial molar volumes (see Fig. 3.8).

This method is more accurate than the method of slopes.

However, the measurement itself need attention and high precision. One have to consider and

analyse the possible error sources of the measurement.

29

3.5 Raoult´s law

The concept of the ideal gas plays an important role in discussions of the thermodynamics of gases and vapors (even if the deviation from ideality is sometimes large).

In case of mixtures it is also useful to define the ideal behaviour. The real systems are characterized by the deviation from ideality.

Ideal gas: complete absence of cohesive forces (Subsection 1.4)

Ideal mixture (liquid, solid): complete uniformity of cohesive forces. If there are two components A and B, the intermolecular forces between A and A, B and B and A and B are all the same.

_mixV = 0, _mixH = 0 _(3.17)

The partial vapor pressure of a component is the measure of the tendency of the component to escape from the liquid phase into the vapor phase.

High vapor pressure means great escaping tendency, and high chemical potential.

The other way round: small chemical potential in liquid phase means small partial pressure in the

vapor phase.

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Raoult´s law: In an ideal liquid mixture the

partial vapor pressure of a component in the vapor phase is proportional to its mole fraction (x) in the liquid phase.

* i i

i

x p

p  

For a pure component x_i = 1, so p_i = p_i*

p_i*: vapor pressure of pure component at the specified temperature.

How does the vapor pressure change with the composition in a two component system?

* 1 1

1

x p

p  

2 2

2

x p

p   p = p

+ p

(3.18)

Since and

(3.19)

1. The p(x) diagram for ideal mixture of two volatile components has the shape like Fig. 3.9.

0 p

x₂  1

t = const.

p p₁*

p₁

p₂*

p₂ ^{Fig. 3.9}

33

2. If only component (1) is volatile like in solutions of solids, the Fig. 3.9 is modified, see Fig. 3.10.

0 p

x₂  1

t = const.

p₁*

p₁= p

In this case the vapor pressure is determined only by

component 1.

Fig. 3.10

Vapor pressure lowering. Based on equations (3.18) and (3.19):

* 1 1

1

x p

p   ^  ^{1  x}

 ^{ p}

 p

 x

 p

* 1

1

* 1

2

p

p x p 



According to (3.20) the relative vapor pressure lowering is equal to the mole fraction of the solute (component 2);

Solute, solvent, solution: see definitions in subsection 3.2.

(3.20)

35

1. Negative deviation: The cohesive forces between unlike molecules are greater than those between the like molecules in pure liquid („like”: the same component).

So the „escaping tendency” is smaller than in ideal

solution. The activity (a) replaces the mole fraction (3.21).

_mixV < 0 ( contraction)

_mixH < 0 ( exotermic solution) p_i < x_i·p_i* p_i = a_i·p_i*

a_i = _i

·x

a_i < x_i

_i < 1

3.6 Deviations from the ideality

(3.21)

The activity coefficient represents the deviation (3.22).

(3.22)

x₂  36

0 1

1 2

p₁*

p₂* t = const.

p

Fig. 3.11 is the isothermal vapor pressure diagram of a two component mixture with negative deviation.

The total vapor pressure may

have a minimum.

Components:

1: chloroform 2: acetone p₁

p₂ p

Fig. 3.11

37

2. Positive deviation: The cohesive forces between unlike molecules are smaller than those between the like molecules in pure liquids („unlike”: from other

component) (see subsection 3.2).

So the „escaping tendency” is greater than in ideal liquid mixture.

_mixV > 0 ( expansion)

_mixH > 0 ( endothermic solution) p_i > x_i·p_i*

a_i > x_i

_i > 1

p_i = a_i·p_i* activity

a_i = _i·x_i

activity coefficient

(3.21, 3.22)

x₂  38

0 1

1 2

p₁*

p₂* t = const.

p

Fig. 3.12 is the isothermal vapor pressure diagram of a two component mixture with positive deviation

The total vapor pressure may

have a maximum.

Components:

1: water 2: dioxane p₁

p₂ p

The deviation character refers

always on the vapor pressure diagram!

Fig. 3.12

39

3.7 Chemical potential in liquid mixtures

1. We derive a formula for calculation of .

2. We use Raoult´s law.

In equilibrium the chemical potential of a component is equal in the two phases (see subsection 2.10).

3.The vapor is regarded as ideal gas (see subsection 1.4).

0

0

ln

p RT p

i g

i

i

    



liq.

vapor

_i

_i^g

* i i

i

x p

p  

0

0

ln

p p RT x

i i

 

 



i i i

i

RT x

p

RT ln p

ln

0





 



Depends on T only: _i* 1. Ideal liquid mixture

i i

i

 

 RT ln x



(3.23)

(3.24)

41

0

0

ln

p RT p

i g

i

i

    



liquid vapor

_i

_i^g

* i i

i

a p

p  

0

0

ln

p p RT a

i i

 

 



i i i

i

RT a

p

RT ln p

ln

0





 



Depends on T only:



*

2. Real liquid mixture

i i

i

 

 RT ln a



i i

i

 

 RT ln a



If x_i  1, _i  1, a_i  1 (pure substance)



*:

_i* = G_mi* (because pure substance).

The activity is a function which replaces the mole fraction in the expression of the chemical potential in case of real solutions.

(3.26)

43

fugacity: “effective” pressure in gas phase activity: “effective” mole fraction

_i(id) deviation from ideality

y_i: mole fraction in gas phase

(3.27)

i i

* i i

i

* i

i











For real gas mixtures the f_i fugacity of the component i

_i is called fugacity coefficient.

0 i

* i 0 i

* i i

i RT ln

p ln p RT p )

ln( f

RT  



     

i i

i p

f    _(3.28)

(3.29)

45

i i

i

 

 RT ln x



Dependence of the chemical potential on the mole fraction (in an ideal liquid mixture)

_i

_i^* x_i 

0 0 1

As the mole fraction approaches zero, the chemical potential

approaches minus infinity (Fig. 3.13)

(3.24)

Fig. 3.13

For most substances the standard chemical potential is negative.

_i

_i^* x_i 

0 0 1

_i* = G_mi* = H_mi* - TS_mi*

Can be

negative or positive

Always positive

(3.30)

Fig. 3.14 introduces the mole fraction dependence in case of negative

chemical potential.

Fig. 3.14

47

Determination of the activity coefficient from liquid- vapor equilibrium data.

Two component system. According to Dalton’s law if the vapor is an ideal gas p_i=y_ip ^(3.31)

vap. y₁,y₂ p₁+p₂ = p liq. x₁,x₂

p₁ = ₁x₁p₁* = y₁p Raoult Dalton

p₂ = ₂x₂p₂* = y₂p

The total pressure and the mole fractions in the liquid and vapor phase are measured. If the vapor pressures of the pure components are known, -s can be

calculated.

* 2 2 2 2

* 1 1 1 1

p x

p y p

x p

y 







(3.32)

3.8 Entropy of mixing and Gibbs free energy of mixing

The quantities of mixing (_mixV, _mixH, _mixS , etc.) are defined at constant temperature and pressure.

We study some important cases:

1. Mixing of ideal gases 2. Ideal mixture of liquids 3. Real mixtures

49

1. Mixing of ideal gases: The two gas components (1 and 2) are separated by a wall, see Fig. 3.15a.

p

Then the wall is

removed. Both gases fill the space (Fig.

3.15b).

There is no interaction (_mixH= 0).

wall

p p

1 2

The pressures of the components are reduced to p₁and p₂: partial

pressures.

p₁= y₁p p₂ = y₂p ^(3.34)

Fig. 3.15a

Fig. 3.15b

Pressure dependence of entropy (at constant temperature):

1

ln

p nR p

S  



the entropy of mixing is the sum of the two entropy changes, see (1.71)

p p R y

p n p R y

n S

S

mix

S

2 2

1 1

2

1

    ln  ln







n₁ = n·y₁ n₂ = n·y₂

 ^y

^{ln y}

^y

^{ln y}



nR

mix

S   



For one mol:





m

mixS  R y ln y  y ln y



(3.35)

(3.36a)

(3.36b)

51

For more than two components:









S ( id ) nR y

ln y

The mole fractions are smaller than 1 so that each term is negative (lny<1).

The entropy of mixing is always positive.

Gibbs function of mixing: _mixG = _mixH - T _mixS 0







G ( id ) nRT y

ln y

It is always negative!

(3.37)

(3.38)

2. Ideal mixture of liquids

First we calculate the Gibbs function of mixing.

n₁

₁*

n₂

₂*

Before mixing:

* 2 2

* 1

)

( initial n  n 

G  

n₁ ₁n₂ ₂ 1

* 1

1

   RT ln x



After mixing:

2

* 2

2

   RT ln x



2 2 1

)

( mixture n  n 

G  

53

2 2

* 2 2 1

1

* 1

1

ln ln

)

( mixture n n RT x n n RT x

G      

2 2

1

1

ln ln

) (

)

( mixture G initial n RT x n RT x G

mix

G    



n₁ = n·x₁ n₂ = n·x₂



^ln

^ln



)

( id nRT x x x x

mix

G  



In case of more than two components:







G ( id ) nRT x

ln x

0

_mixG = _mixH - T _mixS

T S

G

mix

 





It is always positive (the disorder increases by mixing).

We obtained the same expressions for ideal gases and ideal liquid mixtures, compare for entropies (3.37) and (3.40), for Gibbs functions

(3.38) and (3.39), respectively. All equations contain the sums of mole fractions times logarithms of mole fractions.









S ( id ) nR x

ln x

55

kJ/mol

x₂  0

T_mS_m

_mG_m

_mH_m 1.7

-1.7

Ideal mixture:

changes of thermodynamic functions as

functions of mole fraction at about room temperature (Fig. 3.16)

Fig. 3.16

3. Studying the real mixtures, the mole fraction dependence of the thermodynamic

functions _mH_m, T_mS_m and _mG_m depends on the values and signs of the frist two ones.

The subscripts ‘m’ of the -s refer to mixture like in Fig. 3.16.

The next three figures introduce the three possibilities of the relations between the mentioned functions.

For the better understanding of the

properties of mixtures see also subsections 3.1, 3.2 and 3.3.

57

kJ/mol

x₂ 

0

T_mS_m

_mG_m

The entropy of mixing is smaller in real mixtures than in ideal mixtures because there is partial ordering (see subsection 3.2).

_mH_m

There is complete miscibility (compare with Fig. 3.4a).

Real mixture, negative deviation of _mH_m from the ideal behaviour: T_mS_m > _mG_m

Fig. 3.17

58

kJ/mol

x₂ 

0

_mH_m

Case A: _mH_m > T_mS_m,, therefore _mG_m>0.

Therefore the two components are immisible (see Fig.

3.18, compare with Fig. 3.4b)

They are immiscible.

A

_mG_m

If both _mH_m>0 and T_mS_m>0, then two cases are possible

Fig. 3.18

59

kJ/mol

x₂  0

_mG_m

Case B: _mH_m < T_mS_m, therefore

_mG_m<0. The components are

miscible (see Fig. 3.19, compare it with Fig.

3.4a).

B

_mH_m

Fig. 3.19

3.9. Vapor pressure and boiling point diagrams of miscible liquids

Phase rule: F = C - P + 2

In a two component system: F = 4 - P.

In case of one phase there are 3 degrees of freedom.

In case of two phases one parameter has to be kept constant:

Either t = const. (vapor pressure diagram).

Or p = const. (boiling point diagram).

61

Ideal solution See (3.18) and (3.19)

* 1 1

1

x p

p   p

 x

 p

p = p

+p

* 2 2

* 1 1

2 1 2

1

p x

p x

y y p

p



 



* 2

* 1 2

2 2

2

1

1

p p x

x y

y   



< 1

1 1 1 1

2 2





 x

y y ₂ >x ₂

We assume:

*

*

2

1

p

p 

62

First law of Konovalov: y₂ >x₂ i.e. the mole fraction of the more volatile component is higher in the vapor than in the liquid. It is always true when the vapor

pressure does not have a maximum or minimum.

V

Fig. 3.20: vapor pressure diagram, L: liquid curve, V: vapor curve.

Determination of the

vapor pressure diagram:

p p x

p y p

* 2 2

2 2

 



Fig. 3.20

(3.42)

63

In practice the boiling point diagram (temperature-composition diagram) is more important than the vapor pressure diagram (pressure-composition diagram). Distillation at constant pressure is more common than

distillation at constant temperature.

The boiling point of the more volatile component is lower.

64

V: vapor curve

(condensation curve) L: liquid curve

(boiling point curve)

Fig. 3.21

Boiling point diagram

65

Two phase area:

2 2

2

n x n y

x

n 



 



n = n

+ n

2 2

2

2

n x n x n y

x

n















l

x x n y x

n







b n

a

n

 



b a n

n

l

v



The level rule

(3.43)

(3.43): level rule, inverse ratio

x₂,y₂  66

0 1

1 2

p₁*

p₂* t = const.

p

Real solution: Vapor pressure, positive deviation The total vapor pressure may

have a maximum:

azeotrope with maximum

liquid

vapor E.g. Water-dioxane

Water-ethanol

Fig. 3.23

67

Real solution: Boiling point - positive deviation A low-boiling

azeotrope: Fig. 3.24

liquid

vapor

x₂,y₂ 

0 1

1 2

t₁

p = const.

t

t₂

L L

V V

L: boiling point curve

V: condensation curve

E.g. Water-dioxane Water-ethanol

Fig. 3.24

x₂,y₂  68

0 1

1 2

p₁*

p₂* t = const.

p

Real solution:Vapor pressure, negative deviation Total vapor pressure may

have a minimum:

azeotrope with minimum

liquid

vapor

aceton-methanol acetone-chloroform water-nitric acid

Fig. 3.25

69

Real solution: Vapor pressure, negative deviation A high boiling azeotrope.

liquid

vapor

x₂,y₂ 

0 1

1 2

t₁ t

t₂ L L

V V

L: boiling point curve V: condensation curve

acetone-chloroform water-nitric acid

aceton-methanol

Fig. 3.26

1. We start from Gibbs-Duham equation (3.13) (in the liquid phase),

2. The chemical potentials are expressed in terms of vapor pressures

3. The change of total pressure is expressed with respect to mole fraction (dp/dx₂).

3. 10 Thermodynamic

interpretation of azeotropes

71

2

0

2 1

1

d   n d  

n

2

0

2 1

1

d   x d  

x

on composition only.

2 2

2 2

1 1

1

1

dx

d x x dx

d 

 



    



2 2

2 2

1 1

1

1

dx

x x x dx

x 

 

 

  

2

1

dx

dx  

2 2 2

1 1

1

x x

x x



 



  

In equilibrium the chemical potential of a component is equal in the two phases (2.56):

0 0 1

1

1

ln

p RT p



 



ln

p RT p



 



1 1 1

1

ln

x RT p

x 

 



 

2 2 2

2

ln

x RT p

x 

 



 

73

2 2 2

1 1 1

ln ln

x x p

x x p



 





2 2 2

2 1

1 1

1

dx dp p

x dx

dp p

x 

2 1

2

1

1 x dx dx

x    

2 2 2

2 2

1 1

1

dx dp p

x dx

dp p

x 

 

2 2 2

1 2

2 2

1

1 dx

dp p

p x

x dx

dp

 



2 2 2

1 2

2 2

2 2

1

2

1 1

dx dp p

p x

x dx

dp dx

dp dx

dp 



 





 







0

2

2

 dx

dp

We study two cases:

(The total vapor pressure has a maximum or minimum.)

0

2

) ^

dx

A dp

0

2

) ^

dx

B dp

(3.44)

75

0

2

) ^

dx

A dp ₀

1 1

2 1 2

2



 

p p x

x

Dalton: p

= y

p =(1-y

)p p

= y

p

 ¹  ₀

1 1

2 2 2

2

 

 

p y

p y

x

x 1 1

1

2 2

2

 

 y

y x

x

2 2 2

2

1

1

x x y

y  

 1 1

1 1

2 2





 x

y

1 1

2 2

x y

x y





When the total vapor pressure has a

maximum or minimum, the composition of the vapor is equal to the composition of the liquid.

This is the azeotrope point.

Azeotrope is not a compound, the azeotrope composition depends on pressure.

1 1

2 2

x y

x y





(3.45) Third law of Konovalov.